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x^2+0.15x-0.03=0
a = 1; b = 0.15; c = -0.03;
Δ = b2-4ac
Δ = 0.152-4·1·(-0.03)
Δ = 0.1425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.15)-\sqrt{0.1425}}{2*1}=\frac{-0.15-\sqrt{0.1425}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.15)+\sqrt{0.1425}}{2*1}=\frac{-0.15+\sqrt{0.1425}}{2} $
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